Chapter 13 - Surface Areas and Volume, Solved Examples, Class 9, Maths | Extra Documents & Tests for Class 9 PDF Download

Surface Areas and Volumes

INTRODUCTION

Uptill now we have been dealing with plane figures that can be drawn on the page of our notebook or on the blackboard. In this chapter, we shall study about some solid figures like cuboid, cube, cylinder and sphere. We shall also learn to find the surface areas and volumes of these figures.

 CUBOID

A rectangular solid bounded by six rectangular plane faces is called a cuboid. A match box, a tea-packet, a brick, a book, etc.,are all examples of a cuboid.

A cuboid has 6 rectangular faces, 12 edges and 8 vertices.

The following are some definitions of terms related to a cuboid:

(i) The space enclosed by a cuboid is called its volume.

(ii) The line joining opposite corners of a cuboid is called its diagonal. A cuboid has four diagonals.

A diagonal of a cuboid is the length of the longest rod that can be placed in the cuboid.

(iii) The sum of areas of all the six faces of a cuboid is known as its total surface area.

(iv) The four faces which meet the base of a cuboid are called the lateral faces of the cuboid.

(v) The sum of areas of the four walls of a cuboid is called its lateral surface area.

For a cuboid of length = ℓ units, breadth = b units and height = h units, we have :
 

REMARK : For the calculation of surface area, volume etc. of a cuboid, the length, breadth and height must be expressed in the same units.

CUBE

A cuboid whose length, breadth and height are all equal is called a cube.
Ice-cubes, Sugar cubes, Dice, etc. are all examples of a cube. Each edge of a cube is called its side.
For a cube of edge = a units, we have;

CROSS SECTION 

A cut which is made through a solid perpendicular to its length is called its cross section. If the cut has the same shape and size at every point of its length, then it is called uniform cross-section.
Volume of a solid with uniform cross section = (Area of its cross section) × (length).
Lateral Surface Area of a solid with uniform cross section = (Perimeter of cross section) × (length).

SOLVED EXAMPLES

Ex 1. Find the surface area of a cube whose edge is 15 cm.

Sol. The edge of the cube = 15 cm, i.e., a = 15 cm.
Surface area of the cube = 6a² = 6 × (15)² = 1350 cm².

Ex 2. A small indoor greenhouse is made entirely of glass sheets (including the base) held together with tape. It is 40 cm long, 30 cm wide and 30 cm high. Find
 (i) the area of the glass sheet required and
 (ii) the total length of the tape required for all the 12 edges.

Sol. The dimensions of the greenhouse are as under :

Length (ℓ) = 40 cm, Width (b) = 30 cm, Height (h) = 30 cm
The area of the glass sheet required

=  2 [ℓ x b + b x  h + h x  ℓ]
=  2 [40 x 30 + 30 x 30  + 30 x 40] cm²
=  2  [1200 + 900  + 1200] cm²
=  2 x 3300  cm²  =  6600 cm²

Hence, 6600 cm² of glass sheet is required.

Length of the tap required = Sum of the length of the 12 edges.

= 4 (ℓ + b + h)
= 4 x [40 + 30 + 30] cm = 400 cm

Hence, 400 cm of the tape is required.

Ex 3. A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m andthickness of the wall is 24 cm. If this wall is to be built up with bricks of dimensions 24 cm × 12cm × 8 cm, then find the number of bricks which are required.

Sol. We know that, the volume of the wall and the sum of the volumes of the required number of bricks is same.
Length of the wall = 10 × 100 cm = 1000 cm
Breadth or the thickness of the wall = 24 cm
Height of the wall = 4 × 100 cm = 400 cm
The wall is in the shape of a cuboid and its volume = 1000 × 24 × 400 cm³
Now, a brick is also a cuboid having length = 24 cm, breadth = 12 cm and height = 8 cm.
Volume of one brick = 24 × 12 × 8 cm³
The required number of bricks = Volume ofthe wall
Volume of onebrick =  = 4166.6
Hence, the required number of bricks = 4167.

 Ex 4 . Aakriti playing with plastic building blocks which are of identical cubical shapes. She makes a structure as shown in fig. If the edge of each cube is 5 cm, then find the volume of the structure built by Aakriti.

Sol.

In fig. the structure is made with 10 cubes.

Length of an edge of each block = 5 cm.

Volume of one block - (5) 3 cm³ - 125 cm³

Volume of the structure = Volume of the ten blocks = 10 x 125 cm³ = 1250 cm³

Hence, the volume of the structure is 1250 cm³.

RIGHT CIRCULAR CYLINDER

Solids like circular pillars, circular pipes, circular pencils, measuring jars, road rollers and gas cylinders, etc., are said to be in cylindrical shape.

In mathematical terms, a right circular cylinder is a solid generated by the revolution of a rectangle about its sides.

Let the rectangle ABCD revolve about its side AB, so as to describe a right circular cylinder as shown in the figure.

You must have observed that the cross-sections of a right circular cylinder are circles congruent and parallel toeach other.

Cylinders Not Right Circular

There are two cases when the cylinder is not a right circular cylinder.

ln the case-I tap see a cylinder. which is certainly circular but is not at right angles to the base. So we cannot say it is a right circular cylinder.

In the case-II we see a cylinder with a non-circular base as the base is not circular. So we cannot call it a right circular cylinder

REMARK : Unless stated otherwise, here in this chapter the word cylinder would mean a right circular cylinder.

The following are definitions of some terms related to a right circular cylinder :

(i) The radius of any circular end is called the radius of the right circular cylinder.

Thus, in the above figure, AD as well as BC is a radius of the cylinder.

(ii) The line joining the centres of circular ends of the cylinder, is called the axis of the right circular cylinder.

In the above figure, the line AB is the axis of the cylinder. Clearly, the axis is perpendicular to the circular ends.

REMARK : If the line joining the centres of circular ends of a cylinder is not perpendicular to the circular ends, then the cylinder is not a right circular cylinder.

(iii) The length of the axis of the cylinder is called the height or length of the cylinder.

(iv) The curved surface joining the two bases of a right circular cylinder is called its lateral surface.

For a right circular cylinder of radius = r units & height = h units, we have :

The above formulae are applicable to solid cylinders only.

HOLLOW RIGHT CIRCULAR CYLINDERS

Solids like iron pipes, rubber tubes, etc., are in the shape of hollow cylinders.

A solid bounded by two coaxial cylinders of the same height and different radii is called a hollow cylinder.


For a hollow cylinder of height h and with external and internal radii R and r respectively, we have :

Ex 5. A cylindrical block of wood has radius 70 cm and length 2 m is to be painted with blue coloured enamel. The cost of painting is Rs. 1.25 per 100 cm². Find the cost of painting the block. Take π = 22/7

Sol. Here, the radius r of the cylindrical block of wood = 70 cm and the length h = 200 cm
The total surface area of the cylindrical block = 2πr(r + h)
 cm² = 440 × 270 cm² = 118800 cm²
Cost of painting 100 cm² = Rs. 1.25 =

⇒ Cost of painting 1 cm2 = Rs. 5/400
Then, cost of painting 118800 cm² = Rs. 5/400 × 118800 =  × 1188 = Rs.1485
Hence, cost of painting the block of wood = Rs.1485.

Ex 6. A cylindrical vessel, without lid, has to be tin-coated including both of its sides. If the radius ofits base is 1/2 m and its height is 1.4 m, calculate the cost of tin-coating at the rate of Rs. 50 per 1000 cm². (Use π = 3.14)

Sol. Radius of the base (r) = 1/2m =1/2× 100 cm = 50 cm
Height (h) = 1.4 m = 1.4 × 100 cm = 140 cm
Surface area to be tin-coated = 2 (2πrh + πr²)

= 2 [2 x 3.14 x 50 x 140 + 3.14 x (50)²]
= 2 [43960 + 7850] = 2 (51810) = 103620 cm²

= Cost of tin-coating at the rate of Rs. 50 per 1000 cm²

=
Hence, the cost of tin-coating is Rs. 5181.

Ex 7. The pillars of a temple are cylindrical shaped. If each pillar has a circular base of radius 20 cm and height 7 m, then find the quantity of concrete mixture used to build 20 such pillars. Alsofind the cost of the concrete mixture at the rate of Rs. 200 per m³. Take π  = 22/7

Sol. Concrete mixture used for making each pillar = Volume of each pillar = π r² × h

Radius of base of pillar, 

Height of pillar, h = 7m

Volume of mixture used for 1 pillar

Volume of mixture used for 20 pillars

Hence, volume of mixture required to make 20 pillars = 17.6 m³
Now, cost of mixture at the rate of Rs.200 per m³
= Rs.200 × 17.6 = Rs. 3520.
Hence, the cost of concrete mixture is Rs. 3520.

RIGHT CIRCULAR CONE

Solids like an ice-cream cone, a conical tent, a conical vessel, a clown's cap etc. are said to be in conical shape.
In mathematical terms, a right circular cone is a solid generated by revolving a right-angled triangle about one of the sides containing the right angle.
Let a triangle AOC revolve about it's side OC, so as to describe a right circular cone, as shown in the figure.

Cones Not Right Circular

REMARK : Unless stated otherwise, by 'cone' in this chapter, we shall mean 'a right circular cone'.

The following are definitions of some terms related to right circular cone :

(i) The fixed point O is called the vertex of the cone.

(ii) The fixed line OC is called the axis of the cone.

(iii) A right circular cone has a plane end, which is in circular shape. This is called the base of the cone. The vertex of a right circular cone is farthest from its base.

(iv) The length of the line segment joining the vertex to the centre of the base is called the height of the cone. Length OC is the height of the cone.

(v) The length of the line segment joining the vertex to any point on the circular edge of the base, is called the slant height of the cone.
Length OA is slant height of the cone.

(vi) The radius AC of the base circle is called the radius of the cone. There are two cases when we cannot call a cone a right circular cone.

The following are definitions of some terms related to right circular cone :
(i) The fixed point O is called the vertex of the cone.
(ii) The fixed line OC is called the axis of the cone.
(iii) A right circular cone has a plane end, which is in circular shape. This is called the base of the cone. The vertex of a right circular cone is farthest from its base.
(iv) The length of the line segment joining the vertex to the centre of the base is called the height of the cone. Length OC is the height of the cone.
(v) The length of the line segment joining the vertex to any point on the circular edge of the base, is called the slant height of the cone. Length OA is slant height of the cone.
(vi) The radius AC of the base circle is called the radius of the cone.

Relation Between Slant Height, Radius and Vertical Height.

Let us take a right circular cone with vertex at O, vertical height h, slant height ℓ and radius r. A is any point on the rim of the base of the cone and C is the centre of the base. Here, OC = h, AC = r and OA = ℓ.

The cone is right circular and therefore, OC is at right angle to the base of the cone. So, we have OC  CA,
i.e., ΔOCA is right angled at C.
Then by Pythagoras theorem, we have :

For a right circular cone of Radius = r, Height = h & Slant Height = ℓ, we have :

FORMULAE Area of the curved (lateral) surface = πrℓ) sq. units. Total Surface Area of cone = (Curved surface Aiea + Area of Base) = (π rℓ + πr2) sq. units = π r (ℓ + r) sq. units. "Solume of cone

Hollow Right Circular Cone

(i) Centre of the circle is vertex of the cone.
 (ii) Radius of the circle is slant height of the cone.
 (iii) Length of arc AB is the circumference of the base of the cone.
 (iv) Area of the sector is the curved surface area of the cone.

Ex 8. The radius of the base of a conical tent is 12 m. The tent is 9 m high. Find the cost of the canvas required to make the tent, if one square metre of canvas costs Rs. 120. (Take π = 3.14)
 Sol. Here, r = 12 m and h = 9 m. Let  be slant height of the conical tent.

Now, 2 = r² + h² = (12)² + (9)² = 144 + 81 = 225
⇒  = 15 m

The curved surface area of the tent  

Hence, the canvas required = 
Cost of the canvas at the rate of Rs. 120 per m2 = Rs. 120 × 565.2 = Rs. 67824

Ex 9. How many metres of cloth of 1.1 m width will be required to make a conical tent whose vertical height is 12 m and base radius is 16 m? Find also the cost of the cloth used at the rate of Rs.14 per metre.

Sol. Here, h = 12 m, r = 16 m
 = 20m
 Curved surface area =

Width of cloth = 1.1 m

 Cost of the cloth used @ Rs.14 per metre =  × 14 = Rs. 12800.

Ex 10. The curved surface of a right circular cone is 198 cm2 and the radius of its base is 7 cm. Find the volume of the cone. 

Sol. Radius of the base of the cone = 7 cm.
Let h cm be the vertical height and cm be the slant height. Here, r = 7 cm.

Now,   81 – 49 = 32 cm

 × 1.41 cm = 5.64 cm

Hence, the volume of the cone = 289.52 cm³.